1 / 24

Percent Comp

Percent Comp. Percentage composition. Indicates the relative amount of each element present in a compound. Calculating percentage composition. Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each element’s mass by the molar mass.

sheena
Download Presentation

Percent Comp

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Percent Comp

  2. Percentage composition Indicates the relative amount of each element present in a compound.

  3. Calculating percentage composition Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each element’s mass by the molar mass. Step 3: Multiply by 100 to convert to a percentage.

  4. Example 1 1. Calculate the molar mass of water, H2O. Step 1 – H- 2(1.01)= 2.02 O – 1(16.0)= 16.0 ________ 18.02

  5. Step 2 2.02/18.02 X 100 = 11.21% 16.0 /18.02 X 100 = 88.79 % Water is composed of 11.21 % hydrogen and 88.79 % oxygen.

  6. Example 2 Calculate the percentage composition of sucrose , C12H22O11.

  7. Example 3 Find the percentage composition of hydrogen in sulfuric acid.

  8. Empirical Formula From percentage to formula

  9. Empirical Formula • Theempirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxide H2O2 OH Benzene C6H6 CH Ethylene C2H4 CH2 Propane C3H8 C3H8

  10. Calculating Empirical • Pretend that you have a 100 gram sample of the compound. • That is, change the % to grams. • Convert the grams to mols for each element. • Write the number of mols as a subscript in a chemical formula. • Divide each number by the least number. • Multiply the result to get rid of any fractions.

  11. A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H EXAMPLE 1

  12. Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number) EXAMPLE 1

  13. Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C2H5 EXAMPLE 1

  14. Example 2 • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so • 38.67 g C x 1mol C = 3.220 mole C 12.01 gC • 16.22 g H x 1mol H = 16.09 mole H 1.01 gH • 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

  15. Example 2 • 3.220 mole C • 16.09 mole H • 3.219 mole N • C3.22H16.09N3.219 If we divide all of these by the smallest one It will give us the empirical formula

  16. Example 2 • The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N • The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N • C1H5N1 is the empirical formula

  17. Molecular Formula • Is the actual Formula • Only applies to covalent compounds • Find the Mass of the Empirical Formula • Divide the actual molar mass by the mass of one mole of the empirical formula. • Caffeine has a molar mass of 194 g. what is its molecular formula? Empirical Formula C4H5N2O1

  18. Caffeine Example • Find the mass of the empirical formula Empirical Formula C4H5N2O1 C= 4 X 12.01 H = 5 X 1.01 N = 2 X 14.01 O = 1 X 15.99 Total Mass = 97g

  19. Caffeine Example • Find x if • 194 g • 97 g = 2 C4H5N2O1 2 X C8H10N4O2.

  20. Example 2 • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

  21. Example 2 = 2.0mol • 71.65 Cl 24.27C 4.07 H. = 2.0mol = 4.0mol

  22. Example 2 • Cl2C2H4 We divide by lowest (2mol ) • Cl1C1H2 would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

  23. would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? = 2 =

  24. 2 X Cl1C1H2 = Cl2C2H4

More Related